Asked by Thousif
a body is moving in a vertical circle such that the velocities of body at different points are critical the ratio of velocities of body at angular displacements 60° 120° from lowest point is √2:1
Answers
Answered by
Thousif
a body is moving in a vertical circle such that the velocities of body at different points are critical the ratio of velocities of body at angular displacements 60° 120° from lowest point is
And =√2:1
And =√2:1
Answered by
Damon
height h at angle T = R - R cos T
height h at 60 deg = R/2
height h at 120 deg = 3 R/2
kinetic energy
(1/2)mv^2 + m g h = constant
at 60 deg
(1/2)v1^2 + g(R/2)=(1/2)v2^2 + g(3R/2)
(v1^2 -v2^2) = 2 g R
if it just makes it to the top then
(1/2) m v1^2 = m g * 2R
v1^2 = 4 g R
(4 g R - v2^2 ) = 2 g R
v2^2 = 2 g R
v1^2/v2^2 = 4 g R / 2 g R
V1/V2 = sqrt 2 / 1
height h at 60 deg = R/2
height h at 120 deg = 3 R/2
kinetic energy
(1/2)mv^2 + m g h = constant
at 60 deg
(1/2)v1^2 + g(R/2)=(1/2)v2^2 + g(3R/2)
(v1^2 -v2^2) = 2 g R
if it just makes it to the top then
(1/2) m v1^2 = m g * 2R
v1^2 = 4 g R
(4 g R - v2^2 ) = 2 g R
v2^2 = 2 g R
v1^2/v2^2 = 4 g R / 2 g R
V1/V2 = sqrt 2 / 1
Answered by
Anonymous
It's absolutely wrong because we need to find both 60degrees and 120degress
Answered by
Thor
How the hell its wrong!!!
It's perfectly explained!
Anyway thanks for that dude who posted the right one!!!!!
It's perfectly explained!
Anyway thanks for that dude who posted the right one!!!!!
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